Question

if ML$\parallel$BC and NL$\parallel$DC
then prove that $\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}$

Answer 1

Given: ML$\parallel$BC and NL$\parallel$DC

To prove: $\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}$

Proof:

$$\begin{gathered} \mathrm{In}△\mathrm{ABC}, \\ \mathrm{ML}\parallel \mathrm{BC} \\ \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{basic}\mathrm{proportionality}\mathrm{theorem}, \\ \frac{\mathrm{AM}}{\mathrm{MB}}=\frac{\mathrm{AL}}{\mathrm{LC}}...\left(1\right) \\ \mathrm{In}△\mathrm{ADC}, \\ \mathrm{NL}\parallel \mathrm{DC} \\ \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{basic}\mathrm{proportionality}\mathrm{theorem}, \\ \frac{\mathrm{AN}}{\mathrm{ND}}=\frac{\mathrm{AL}}{\mathrm{LC}}...\left(2\right) \\ \mathrm{From}\left(1\right)\&\left(2\right),\mathrm{we}\mathrm{get}: \\ \frac{\mathrm{AM}}{\mathrm{MB}}=\frac{\mathrm{AN}}{\mathrm{ND}} \\ \mathrm{By}\mathrm{invertendo},\mathrm{we}\mathrm{have}: \\ \frac{\mathrm{MB}}{\mathrm{AM}}=\frac{\mathrm{ND}}{\mathrm{AN}} \\ \mathrm{Adding}1\mathrm{on}\mathrm{both}\mathrm{the}\mathrm{sides},\mathrm{we}\mathrm{get}: \\ \frac{\mathrm{MB}}{\mathrm{AM}}+1=\frac{\mathrm{ND}}{\mathrm{AN}}+1 \\ \Rightarrow \frac{\mathrm{MB}+\mathrm{AM}}{\mathrm{AM}}=\frac{\mathrm{ND}+\mathrm{AN}}{\mathrm{AN}} \\ \Rightarrow \frac{\mathrm{AB}}{\mathrm{AM}}=\frac{\mathrm{AD}}{\mathrm{AN}} \\ \mathrm{By}\mathrm{invertendo},\mathrm{we}\mathrm{have}: \\ \frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}} \end{gathered}$$